Question

решите уровнение
sin(2x-π)-1=0

Answer 1

$sin(2x-\pi)-1=0 \\ sin(2x-\pi)=1 \\ 2x-\pi= \frac{\pi}{2}+2\pi n \\ 2x= \frac{3\pi}{2}+2\pi n \\ x= \frac{3\pi}{4}+\pi n, n\in Z$